A) \[nR\]
B) \[{{n}^{2}}R\]
C) \[\frac{n}{R}\]
D) \[\frac{R}{{{n}^{2}}}\]
Correct Answer: B
Solution :
When wire is stretched, its volume remains same, hence\[lA=\]constant where \[l\] is length and A the area. \[{{l}_{1}}{{A}_{1}}={{l}_{2}}{{A}_{2}}\] \[\Rightarrow \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] Given, \[{{l}_{2}}=n\,\,{{l}_{1}}\] \[\therefore \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{1}{n}\] The resistance of a wire of length \[l\], area A and specific resistance p is \[R=\rho \frac{l}{A}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}\] Given, \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{n}\], \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{1}{n}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{{{n}^{2}}}\] \[\Rightarrow \] \[{{R}_{2}}={{n}^{2}}{{R}_{1}}\] Given, \[{{R}_{1}}=R\] \[\therefore \] \[{{R}_{2}}={{n}^{2}}R\]
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